= Solution
For a <cycle type> $\mu=(m,\nu)$, the <Murnaghan–Nakayama rule> is
$$
\chi^\lambda(m,\nu)=\sum_{\lambda/\rho\text{ a rim hook of size }m}
(-1)^{\operatorname{ht}(\lambda/\rho)}\chi^\rho(\nu).
$$
The sum is over partitions $\rho$ whose diagrams remain after removing the specified border strip; $\nu$ is the remaining <cycle type>, and the recursion ends with the empty <character> equal to one. At the identity, $\chi^\rho(1^{|\rho|})=f^\rho$. Incompatible total sizes contribute zero.
For shape $(2,2)$ and a transposition, a horizontal strip in the bottom row leaves $(2)$ with positive sign; a vertical strip in the right column leaves $(1,1)$ with negative sign. Both remaining dimensions are one. Thus
$$
\boxed{\chi^{(2,2)}(2,1,1)=1-1=0.}
$$
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